If the subspace $A$ of a metric space $B$ is homeomorphic to a metric space $C$, then $C$ is the subspace of a metric space homeomorphic to $B$ In my proof to an exercice, I used this lemma :

If the subspace $A$ of a metric space $B$ is homeomorphic to a metric space $C$, then $C$ is the subspace of a metric space homeomorphic to $B$

I think it's true and I think I've proved it, but I want to double check and couldn't find anything on the internet...
My proof goes roughly like this : We can add elements to $C$ so that we get a topological space $D$ homeomorphic to $B$. But since $B$ is metrizable, so is $D$.
Does it hold?
Thank you for your time.
 A: Your "proof" is not a proof. It's the idea of a proof. It's a good idea, and worth pursuing further, but so far, it's just that. To actually prove the statement, you need to


*

*Define $D$.

*Define the metric on $D$.

*Prove that the metric, applied on $C$, is the same as the original metric on $C$ (depending on how you perform point 2, this part may be trivial).

*Prove that $D$ is homeomorphic to $B$. To do that, you will have to define a homeomorphism.


Hint: maybe try to construct $D$ simply by taking $C$, and adding to it all elements that are in $B$ but not in $A$. 
A: Edit: The counterexample below makes assumptions that haven't been specified in the question: that the desired extension commutes with the homeomorphisms.
See here for a proper answer (and question!).

The answer is No. In fact the lemma is plain wrong and here is a counterexample:
Take $B$ a converging sequence and $A$ the sequence  minus the limit point. Then we can have $C$ a discrete space. $C$ can't be extended to get something homeomorphic to $B$
